3.64 \(\int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=72 \[ -\frac {2 b \log (\tan (c+d x))}{a^3 d}+\frac {2 b \log (a+b \tan (c+d x))}{a^3 d}-\frac {b}{a^2 d (a+b \tan (c+d x))}-\frac {\cot (c+d x)}{a^2 d} \]

[Out]

-cot(d*x+c)/a^2/d-2*b*ln(tan(d*x+c))/a^3/d+2*b*ln(a+b*tan(d*x+c))/a^3/d-b/a^2/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3516, 44} \[ -\frac {b}{a^2 d (a+b \tan (c+d x))}-\frac {2 b \log (\tan (c+d x))}{a^3 d}+\frac {2 b \log (a+b \tan (c+d x))}{a^3 d}-\frac {\cot (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2/(a + b*Tan[c + d*x])^2,x]

[Out]

-(Cot[c + d*x]/(a^2*d)) - (2*b*Log[Tan[c + d*x]])/(a^3*d) + (2*b*Log[a + b*Tan[c + d*x]])/(a^3*d) - b/(a^2*d*(
a + b*Tan[c + d*x]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps

\begin {align*} \int \frac {\csc ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {1}{x^2 (a+x)^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac {b \operatorname {Subst}\left (\int \left (\frac {1}{a^2 x^2}-\frac {2}{a^3 x}+\frac {1}{a^2 (a+x)^2}+\frac {2}{a^3 (a+x)}\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac {\cot (c+d x)}{a^2 d}-\frac {2 b \log (\tan (c+d x))}{a^3 d}+\frac {2 b \log (a+b \tan (c+d x))}{a^3 d}-\frac {b}{a^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.40, size = 109, normalized size = 1.51 \[ \frac {-a^2 \cot ^2(c+d x)+b^2 (2 \log (a \cos (c+d x)+b \sin (c+d x))-2 \log (\sin (c+d x))+1)-a b \cot (c+d x) (-2 \log (a \cos (c+d x)+b \sin (c+d x))+2 \log (\sin (c+d x))+1)}{a^3 d (a \cot (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2/(a + b*Tan[c + d*x])^2,x]

[Out]

(-(a^2*Cot[c + d*x]^2) - a*b*Cot[c + d*x]*(1 + 2*Log[Sin[c + d*x]] - 2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]) +
b^2*(1 - 2*Log[Sin[c + d*x]] + 2*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]))/(a^3*d*(b + a*Cot[c + d*x]))

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fricas [B]  time = 0.46, size = 293, normalized size = 4.07 \[ -\frac {a^{2} b^{2} - {\left (a^{4} + 2 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} b^{2} + b^{4} - {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - {\left (a^{2} b^{2} + b^{4} - {\left (a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right )}{{\left (a^{5} b + a^{3} b^{3}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{6} + a^{4} b^{2}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - {\left (a^{5} b + a^{3} b^{3}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-(a^2*b^2 - (a^4 + 2*a^2*b^2)*cos(d*x + c)^2 - (a^3*b + 2*a*b^3)*cos(d*x + c)*sin(d*x + c) + (a^2*b^2 + b^4 -
(a^2*b^2 + b^4)*cos(d*x + c)^2 + (a^3*b + a*b^3)*cos(d*x + c)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c
) + (a^2 - b^2)*cos(d*x + c)^2 + b^2) - (a^2*b^2 + b^4 - (a^2*b^2 + b^4)*cos(d*x + c)^2 + (a^3*b + a*b^3)*cos(
d*x + c)*sin(d*x + c))*log(-1/4*cos(d*x + c)^2 + 1/4))/((a^5*b + a^3*b^3)*d*cos(d*x + c)^2 - (a^6 + a^4*b^2)*d
*cos(d*x + c)*sin(d*x + c) - (a^5*b + a^3*b^3)*d)

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giac [A]  time = 0.89, size = 74, normalized size = 1.03 \[ \frac {\frac {2 \, b \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{3}} - \frac {2 \, b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {2 \, b \tan \left (d x + c\right ) + a}{{\left (b \tan \left (d x + c\right )^{2} + a \tan \left (d x + c\right )\right )} a^{2}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

(2*b*log(abs(b*tan(d*x + c) + a))/a^3 - 2*b*log(abs(tan(d*x + c)))/a^3 - (2*b*tan(d*x + c) + a)/((b*tan(d*x +
c)^2 + a*tan(d*x + c))*a^2))/d

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maple [A]  time = 0.47, size = 75, normalized size = 1.04 \[ -\frac {b}{a^{2} d \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{3} d}-\frac {1}{d \,a^{2} \tan \left (d x +c \right )}-\frac {2 b \ln \left (\tan \left (d x +c \right )\right )}{a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x)

[Out]

-b/a^2/d/(a+b*tan(d*x+c))+2*b*ln(a+b*tan(d*x+c))/a^3/d-1/d/a^2/tan(d*x+c)-2*b*ln(tan(d*x+c))/a^3/d

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maxima [A]  time = 0.57, size = 74, normalized size = 1.03 \[ -\frac {\frac {2 \, b \tan \left (d x + c\right ) + a}{a^{2} b \tan \left (d x + c\right )^{2} + a^{3} \tan \left (d x + c\right )} - \frac {2 \, b \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{3}} + \frac {2 \, b \log \left (\tan \left (d x + c\right )\right )}{a^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-((2*b*tan(d*x + c) + a)/(a^2*b*tan(d*x + c)^2 + a^3*tan(d*x + c)) - 2*b*log(b*tan(d*x + c) + a)/a^3 + 2*b*log
(tan(d*x + c))/a^3)/d

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mupad [B]  time = 3.84, size = 79, normalized size = 1.10 \[ \frac {2\,b\,\ln \left (\frac {a+b\,\mathrm {tan}\left (c+d\,x\right )}{\mathrm {tan}\left (c+d\,x\right )}\right )}{a^3\,d}-\frac {2\,b}{a^2\,d\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {1}{a\,d\,\mathrm {tan}\left (c+d\,x\right )\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(c + d*x)^2*(a + b*tan(c + d*x))^2),x)

[Out]

(2*b*log((a + b*tan(c + d*x))/tan(c + d*x)))/(a^3*d) - (2*b)/(a^2*d*(a + b*tan(c + d*x))) - 1/(a*d*tan(c + d*x
)*(a + b*tan(c + d*x)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2/(a+b*tan(d*x+c))**2,x)

[Out]

Integral(csc(c + d*x)**2/(a + b*tan(c + d*x))**2, x)

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